# Area models for multiplication throughout the K-12 curriculum

Let’s take a look at area models, shall we?

My thesis today is that area models should be ubiquitous across the entire curriculum because mathematics is a sense making discipline. As math educators, we ought to encourage our students to take every opportunity to visualize their mathematics in an effort to illuminate, explain, prove, and bring intuition.

So let’s take a walk through the K-12 math curriculum and highlight the use of area models as they might apply to arithmeticalgebra, and calculus.

# Arithmetic

Students experience area models for the first time in elementary school as they work to visualize multi-digit multiplication. This can also be used for division as well, just running the logic in reverse–that is, seeking an unknown “side length” rather than an unknown area. And Base Ten Blocks can be used to help students understand the building blocks of our number system.

Here’s how you might work out $27\times 54$:

$27\times 54 = (20+7)(50+4)=(20)(50)+(20)(4)+(7)(50)+(7)(4)$

$27\times 54=1000+80+350+28=1458$

The advantage of using a visual model like this is that you can easily see your calculation and explain why constituent calculations, taken together, faithfully produce the desired result. If you do a “man on the street” interview with most users or purveyors of the standard algorithm, you would almost certainly not get crystal clear explanations for why it produces results. For a further discussion of area models for multi-digit multiplication, see this article, or read Jo Boaler’s now famous book Mathematical Mindsets.

# Algebra

In middle school, as students first encounter algebra, they may use area models to support their algebraic reasoning around multiplying polynomials. And in an Algebra 2 course they may learn about polynomial division and support their thinking using an area model in the same way they used area models to do division in elementary school. Here Algebra Tiles can be used as physical manipulatives to support student learning.

Here’s how you might work out $(x+4)(2x+3)$:

$(x+4)(2x+3)=(x)(2x)+(x)(3)+(4)(2x)+(4)(3)$

$(x+4)(2x+3)=2x^2+3x+8x+12=2x^2+11x+12$

Notice also that if you let $x=10$, you obtain the following result from arithmetic:

$14\times 23 = 200+110+12=322$

The Common Core places special emphasis on making such connections. I agree with this effort, even though I can also commiserate with fellow math teachers who say things like, “My Precalculus students still use the box method for multiplying polynomials!” We definitely want to move our students toward fluency, but perhaps we should wait for them to realize that they don’t need their visual models. Eventually most students figure out on their own that it would be more efficient to do without the models.

# Calculus

Later in high school, as students first study calculus, area models can be used to bring understanding to the Product Rule–a result that is often memorized without any understanding. Even the usual “textbook proof” justifies but does not illuminate.

Here’s an informal proof of the Product Rule using an area model:

The “change in” the quantity $L\cdot W$ can be thought of as the change in the area of a rectangle with side lengths $L$ and $W$. That is, let $A=LW$. As we change $L$ and $W$ by amounts $\Delta L$ and $\Delta W$, we are wondering how the overall area changes (that is, what is $\Delta A$?).

If the side length $L$ increases by $\Delta L$, the new side length is $L+\Delta L$. Similarly, the width is now $W+\Delta W$. It follows that the new area is:

$A+\Delta A=(L+\Delta L)(W+\Delta W)=LW+L\Delta W+W\Delta L+\Delta L\Delta W$

Keeping in mind that $A=LW$, we can subtract this quantity from both sides to obtain:

$\Delta A=L\Delta W+W\Delta L+\Delta L\Delta W$

Dividing through by $\Delta x$ gives:

$\frac{\Delta A}{\Delta x}=L\cdot\frac{\Delta W}{\Delta x}+W\cdot\frac{\Delta L}{\Delta x}+\frac{\Delta L}{\Delta x} \frac{\Delta W}{\Delta x} \Delta x$

And taking limits as $\Delta x\to 0$ gives the desired result:

$\frac{dA}{dx}=L\cdot\frac{dW}{dx}+W\cdot\frac{dL}{dx}$

# Conclusion

If you’re like me, you once looked down on area models as being for those who can’t handle the “real” algebra. But if we take that view, there’s a lot of sense-making that we’re missing out on. Area models are an important tool in our tool belt for bringing clarity and connections to our math students.

Okay, so last question: Base Ten Blocks exist, and Algebra Tiles exist. What do you think? Shall we manufacture and sell Calculus DX Tiles © ? 🙂

# Extraneous Solutions – Part 1 of 3?

## Disclaimer

Within my small inner circle of math teachers, the mystery of extraneous solutions seems to be the issue of the year. I have so much to say on this topic (algebraic, logical, pedagogical, historical, linguistic) that I don’t really know where to begin. My only disclaimer is that I’m not really sure if this topic is all that important.

## Solving an Equation with a Radical Expression

Consider the following equation:

(1) $2\sqrt{x+8} +5 = 11$

One hardly needs algebra skills or prior knowledge to solve this, but prior experience suggests trying to isolate $x$.

(2) $2\sqrt{x+8} = 6$ (we subtract 5 from both sides)

(3) $\sqrt{x+8} = 3$ (we divide both sides by 2)

Now, if the square root of something is 3, then that something must be 9, so it immediately follows that

(4) $x+8 = 9$

(5) $x = 1$ (we subtract 8 from both sides)

## Squaring Both Sides

In my transition from (3) to (4), I used a bit of reasoning. Some conversational common sense told me that “if the square root of something is 3, then that something must be 9”. But that logic is usually just reduced to an algebraic procedure: “squaring both sides”. If we square both sides of equation (3), we get equation (4).

On the one hand, this seems like a natural move. Since the meaning of $\sqrt{a}$ is “the (positive) quantity which when squared is $a$“, the expression $\sqrt{a}$ is practically begging us to square it. Only then can we recover what lies inside. A quantity “which when squared is $a$” is like a genie “which when summoned will grant three wishes”. In both cases you know exactly what to do next.

Unfortunately, squaring both sides of an equation is problematic. If $a = b$ is true, then $a^2 = b^2$ is also true. But the converse does not hold. If $a^2 = b^2$, we cannot conclude that $a = b$, because opposites have the same square.

This leads to problems when solving an equation if one squares both sides indiscriminately.

## A Silly Equation Leads to Extraneous Solutions

Consider the equation,

(6) $x = 4$

This is an equation with one free variable. It’s a statement, but it’s a statement whose truth is impossible to determine. So it’s not quite a proposition. Logicians would call it a predicate. Linguistically, it’s comparable to a sentence with an unresolved anaphor. If someone begins a conversation with the sentence “He is 4 years old”, then without context we can’t process it. Depending on who “he” refers to, the sentence may be true or false. The goal of solving an equation is to find the solution set, the set of all values for the free variable(s) which make the sentence true.

Equation (6) is only true if $x$ has value 4. So the solution set is $\left\{4 \right\}$. But if we square both sides for some reason…

(7) $x^2 = 16$ has solution set $\left\{4, -4\right\}$

We began with $x = 4$, “did some algebra”, and ended up with $x^2 = 16$. By inspection, $-4$ is a solution to $x^2 = 16$, but not to the original equation which we were solving, so we call $-4$ an “extraneous solution”. [Extraneous – irrelevant or unrelated to the subject being dealt with]

Note that the appearance of the extraneous solution in the algebra of (6)-(7) did not involve the square root operation at all. But this example was also a bit silly because no one would square both sides when presented with equation (6), so let’s look at a slightly less silly example.

(8) $2\sqrt{x+8} + 5 = -1$

(9) $2\sqrt{x+8} = -6$

(10) $\sqrt{x+8} = -3$

People paying attention might stop here and conclude (correctly) that (10) has no solutions, since the square root of a number can not be negative. Closer inspection of the logic of the algebraic operations in (8)-(10) enables us to conclude that the original equation (8) has no solutions either. Since $a = b \iff a - 5 = b -5$, any solution to (8) will also be a solution to (9) and vice versa. Since $a = b \iff a/2 = b/2$, any solution to (9) will also be a solution to (10) and vice versa. So equations (8), (9), and (10) are all “equivalent” in the sense that they have the same solution set.

But what if the equation solver does not notice this fact about (10) and decides to square both sides to get at that information hidden inside the square root?

(11) $x+8 = 9$

(12) $x = 1$

Again we have an extraneous solution. $x = 1$ is a solution to (12), but not to the original equation (8). Where did everything go wrong? By the previous logic, (8), (9), and (10) are all equivalent. (11) and (12) are also equivalent. So the extraneous solution somehow arose in the transition from (10) to (11), by squaring both sides.

So unlike subtracting 5 from both sides or dividing both sides by 2, squaring both sides is not an equivalence-preserving operation. But we tolerate this operation because the implication goes in the direction that matters. If $a = b$, then $a^2 = b^2$, so if $a$ and $b$ are expressions containing a free variable $x$, any value of $x$ that makes $a = b$ true will also make $a^2 = b^2$ true.

In other words, squaring both sides can only enlarge the solution set. So if one is vigilant when squaring both sides to the possible creation of extraneous solutions, and is willing to test solutions to the terminal equation back into the original equation, the process of squaring both sides is innocent and unproblematic.

## Those Who are Still Not Satisfied

Still there are some who are not satisfied with this explanation: “Why does this happen? What is really going on? Where do the extraneous solutions come from? What do they mean?”

One source of the problem is the square root operation itself. $\sqrt{a}$ is, by the conventional definition, the positive quantity which when squared is $a$. The reason that we have to stress the positive quantity is that there are always two real numbers that when squared equal any given positive real number. There are a few slightly different ways of making this same point. The operation of squaring a number erases the evidence of whether that number was positive or negative, so information is lost and we are not able to reverse the squaring process.

We can also phrase the phenomenon in the language of functions. Since squaring is a common and useful mathematical practice, information will often come to us squared and we’ll need an un-squaring process to unpack that information. $f(x) = x^2$, for all the reasons just mentioned, is not a one-to-one function, so strictly speaking, it is not invertible. But un-squaring is too important, so we persevere. As with all non-one-to-one functions, we first restrict the domain of $f(x) = x^2$ to $[0, \infty)$ to make it one-to-one. This inverse, $f^{-1}(x) = \sqrt{x}$ thus has a positive range and so the convention that $\sqrt{a} \geq 0$ is born. So every use of the square root symbol comes with the proviso that we mean the positive root, not the negative root. We inevitably lose track of this information when squaring both sides.

[Note: Students can easily lose track of these conventions. After a lot of practice solving quadratic equations, moving from $x^2 = 9$ effortlessly to $x = \pm 3$, students will often start to report that $\sqrt{9} = \pm 3$.]

The convention that we choose the positive root is totally arbitrary. In a world in which we restricted the domain of  $f(x) = x^2$ to $(-\infty, 0]$ before inverting, $\sqrt{9}$ would be $-3$. In that world, $x = 1$ is a perfectly good solution to $2\sqrt{x+8} + 5 = -1$, not extraneous at all.

## A Trigonometric Equation which Yields an Extraneous Solution

For parallelism, consider the (somewhat artificial) equation:

(13) $\arccos(2x-1) = \frac{4\pi}{3}$

Like in (10), careful and observant solvers might notice that the range of the $\arccos(x)$ function is $[0, \pi]$ and correctly conclude that the equation has no solutions. But there seems to be a lot going on inside that $\arccos$ expression, so many will rush ahead and try to unpack it by “cosineing”. Indeed, since $a=b \Rightarrow \cos(a) = \cos(b)$, this seems innocent.

(14) $2x - 1 = -\frac{1}{2}$

(15) $2x = \frac{1}{2}$

(16) $x = \frac{1}{4}$

But $x = \frac{1}{4}$ is an extraneous solution since $\arccos(-\frac{1}{2}) = \frac{2\pi}{3}$ not $\frac{4\pi}{3}$.

The explanation for this extraneous solution will be similar to the logic we used above. If $a = b$, then $\cos(a) = \cos(b)$, so if $a$ and $b$ are expressions containing a free variable $x$, any value of $x$ that makes $a = b$ true will also make $\cos(a) = \cos(b)$ true. So we will not lose any solutions by “taking the cosine of both sides”. But as the cosine function is not one-to-one, $\cos(a) = \cos(b)$ does not imply that $a = b$. So taking the cosine of both sides, just like squaring both sides, can enlarge the solution set.

The above paragraph explains why extraneous solutions could appear in the solution of (13), but maybe not why they do appear. For that, we again must look to the presence of the $\arccos$ function. Since $\cos$ is not one-to-one, we had to arbitrarily restrict its domain to $[0, \pi]$ prior to inverting. So every use of the $\arccos$ symbol comes with its own proviso that we are referring to a number in a particular interval of values. In a world in which we had restricted the domain of $\cos$ to $[\pi, 2\pi]$ prior to inverting, $x = \frac{1}{4}$ would be a perfectly good solution to $\arccos(2x-1) = \frac{4\pi}{3}$, not extraneous at all.

The above examples seem to suggest that one can avoid dealing with extraneous solutions by carefully examining one’s equations at each step. But in practice, this really isn’t possible. I saved the fun examples for the end, but as this post is already way way too long, they will have to wait for a bit later.

-Will Rose

## Thanks

Thanks to John Chase for letting me guest post on his blog. Thanks to James Key for encouraging me again and again to think about extraneous solutions.

# Half-your-age-plus-seven rule

Looking for a great application of systems of linear inequalities for your Algebra 1 or 2 class? Look no further than today’s GraphJam contribution:

You might just give this picture to students and ask THEM to come up with the equations of the three lines.

There’s also a nice discussion to be had here about inverse functions, or about intersecting lines. And there might also be a good discussion about the domain of reasonableness.

Here are the three functions:

$f_{\text{blue}}(x)=x$

$f_{\text{red}}(x)=\frac{1}{2}x+7$

$f_{\text{black}}(x)=2x-14$

This is especially interesting because I never think of the rule as putting boundaries on a person’s dating age range. Usually people talk about it in the context of “how old of a person can I date?” not “how young of a person can I date?” Or rather, if you’re asking the second question, it’s usually phrased “how young of a person can date me?” (All of these questions relate to functions and their inverses!) But in fact, the half-your-age-plus-seven rule puts a lower and and upper bound on the ages of those you can date.

As far as reasonableness, is it fair to say that my daughter who is 1 can date someone who is between the age of -12 and 7.5? I don’t think so! I’m definitely going to be chasing off those -12 year-olds, I can already tell :-).

For my daughter, the domain of reasonableness might be $x\geq 18$!

# Pictures with equations

Check out this awesome blog post by Richard Clark on the Alpha Blog.

Follow the link to see lots of great pictures made with equations. These pictures are so complicated it makes you wonder, is there any picture we can’t make with equations? My first answer is NO.

Think about vector-based graphics. Vector graphics, for those who’ve never heard the term, are pictures/graphics that are stored as a set of instructions for redrawing the picture rather than as a large array of pixels. You’ve used vector graphics if you have ever used clip-art or used the drawing tools in Microsoft Office, or if you’ve ever used Adobe Illustrator, or Inkscape. The advantages of vector graphics include very small files and infinite loss-less resizeability. How can vector graphics achieve this? Well, like I said, vector graphics are stored as rules not pixels. And by rules, we could just as easily say equations.

So the answer is certainly YES we can make any picture using equations. I think the harder question is can we make any picture using ONE equation? Or one set of parametric equations? Or one implicit equation?

What constraints do we want to impose? Do fractals/iterative/recursive rules count?

I am curious to find out how the creators of these picture-equations came up with them. It seems infeasible to do this by trial and error, given the massive size of these equations.

Oh, and if you haven’t yet seen the Batman Curve, you better go check that out too.

# “Japanese” Multiplication

My brother sent me a link to this video that teaches “Japanese” multiplication (thanks Tim!):

I learned about this technique in my History of Math class, and Vi Hart talked about it in a video back in 2011:

She does a nice job showing why there’s nothing particularly special about this Japanese “visual” multiplication. Here are a few reasons why it’s not better, as far as I’m concerned:

1. It’s not faster (sometimes it is, but most of the time not). As Vi points out, counting the number of dots in a rectangle by hand is ridiculous.
2. It’s painful when the numbers are bigger than 1, 2, or 3 and when there are more than 2 digits in the numbers (just try multiplying 976 x 8937 for example).
3. Zeros make things difficult (use dashed lines?)
4. Carrying is still required.
5. It’s perhaps more error prone, since it relies on your counting all the intersections.

In the end, to multiply two numbers you still have to multiply all their digits by each other and deal with carries, no matter which method you choose. I think it’s still worth teaching various methods of multiplication to students in an effort to make the abstract more concrete.

# Why does x represent the unknown?

I think I’ll show this to my Algebra 2 class this week.

[HT: Fred Connington]

# Rationalization Rant

Every high school math student has been taught how to rationalize the denominator. We tell students not to give an answer like

$\frac{1}{\sqrt{2}}$

because it isn’t fully “simplified.” Rather, they should report it as

$\frac{\sqrt{2}}{2}.$

This is fair, even though the second answer isn’t much simpler than the first. What does it really mean to simplify an expression? It’s a pretty nebulous instruction.

We also don’t consider

$\frac{12}{1+\sqrt{5}}$

to be rationalized because of the square root in the denominator, so we multiply by the conjugate to obtain

$2-2\sqrt{5}.$

In this particular example, multiplying by the conjugate was really fruitful and the resulting expression does indeed seem much more desirable than the original expression.

But here’s where it gets a little ridiculous. Our Algebra 2 book also calls for students to rationalize the denominator when (1) a higher root is present and (2) roots containing variables are present. Let me show you an example of each situation, and explain why this is going a little too far.

## Rationalizing higher roots

First, when a higher root is present like

$\sqrt[5]{\frac{15}{2}},$

the book would have students multiply the top and bottom of the fraction inside the radical by $2^4$ so as to make a perfect fifth root in the denominator. The final answer would be

$\frac{\sqrt[5]{240}}{2}.$

Simpler? You decide.

This becomes especially problematic when we encounter sums involving higher roots. It’s certainly possible, using various tricks, to rationalize the denominator in expressions like this:

$\frac{1}{2-\sqrt[3]{5}}.$

But is that really desirable? The result here is

$\frac{1}{2-\sqrt[3]{5}}\cdot\frac{4+2\sqrt[3]{5}+\sqrt[3]{25}}{4+2\sqrt[3]{5}+\sqrt[3]{25}}=\frac{4+2\sqrt[3]{5}+\sqrt[3]{25}}{3},$

which is, arguably, more complex than the original expression. Can anyone think of a good reason to do this, except just for fun?

## Rationalizing variable expressions

Now, let’s think about variable expressions. Here is a problem, directly from our Algebra 2 book (note the directions as well):

Write the expression in simplest form. Assume all variables are positive.

$\sqrt[3]{\frac{x}{y^7}}$

The method that leads to the “correct” solution is to multiply the fraction under the radical by $\frac{y^2}{y^2}$, and to finally write

$\frac{\sqrt[3]{xy^2}}{y^3}.$

This is problematic for two reasons. (1) This isn’t really simpler than the original expression and (2) this expression isn’t even guaranteed to have a denominator that’s rational! (Suppose $y=\sqrt{2}$ or even $y=\pi$.) Once again I ask, can anyone think of a good reason to do this, except just for fun??

## So how far do we take this?

Is it reasonable to ask someone to rationalize this denominator?

$\frac{1}{2\sqrt{2}-\sqrt{2}\sqrt[3]{5}+2\sqrt{5}-5^{5/6}}$

You can rationalize the denominator, but I’ll leave that as an exercise for the reader. So how far do we take this? I had to craft the above expression very carefully so that it works out well, but in general, most expressions have denominators that can’t be rationalized (and I do mean “most expressions” in the technical, mathematical way–there are are an uncountable number of denominators of the unrationalizable type). All that being said, I think this would make a great t-shirt:

And I rest my case.